Get an old L.E.D. and put a load on the resistor, if I recall correctly from my days in the electronics field, there has to be a load to get a voltage reading and the resistor has to be "out of the circuit" completely to get the resistance reading.
A 1,000ohm{1K} resistor is good, but a better alternative would be an LM7805 Voltage Regulator, as long as the + output is DC and always +, then the LM7805 will give a constant 5VDC output.
You'd wire the LM7805 {+}input pin to the + side of the + output of the controlling device, the center pin goes to ground or the - side, the 3rd pin is the output pin {+ 5VDC} and that would go to the + leg of the L.E.D.'s it would be powering.
One thing to note, the LM7805 can get quite hot, so a heatsink would be recommended for it. Also the tab of the LM7805 has a hole for a screw, but be careful as this metal tab is linked internally to the middle pin {-} or ground of the LM7805.
And to answer your question if the resistor is NOT of a high enough wattage to handle the current, yes, you can burn it out. Believe it or not the heating elements in some smoke units, the ones that can be replaced in, for example, fan driven smoke units, are actually resistors and these burn out, they just an inside out resistor to put it simpley, but the correct name for them is actually "thermistor" because the voltage is what causes them to heat, the higher the voltage the hotter they get, too hot and they burn out. Same with a standard resistor.
Most G loco's usually need at a bare minimum, depending is 1/4 watt, you really shouldn't need over a 1/2 watt resistor for most lighting applications. Minimum resistance I've used is 470 ohm, but most often I up that to 1k {1,000 ohm}, it does diminish the L.E.D. brightness, but not enough to really notice, especially if you use "super bright" L.E.D. types.
Good Luck!